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anonymous
 one year ago
if Given
E=mc^2
C representing the constant Distance/Time.
What if Time is then considered to actually be a single local computation cycle. But the computational cycle is not constant throughout the universe, and is made longer by computational complexity. So modified by the variable x say, which represents an extension to the clock cycle, caused by compleXity. Local space then is something like a fixed clock computer, that gets slowed down by extra jobs added to it, thus slowing all its jobs. And we get E=m(d/(t+x))^2. Now if x increases, the other variables must change. Any thoughts?
anonymous
 one year ago
if Given E=mc^2 C representing the constant Distance/Time. What if Time is then considered to actually be a single local computation cycle. But the computational cycle is not constant throughout the universe, and is made longer by computational complexity. So modified by the variable x say, which represents an extension to the clock cycle, caused by compleXity. Local space then is something like a fixed clock computer, that gets slowed down by extra jobs added to it, thus slowing all its jobs. And we get E=m(d/(t+x))^2. Now if x increases, the other variables must change. Any thoughts?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could this equation describe what we observe?

perl
 one year ago
Best ResponseYou've already chosen the best response.2C represents the speed of light, its a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0true, but Im saying, what if we changed that. and said.. it isn't constant.. and we modified it.. to include an interaction between itself and everything that is going on around it.. Took away some computational energy to deal with an increase in information exchange and complexity. And then compare that to what we observe when it comes to black holes, and universal expansion. I was just curious to see what people a lot smarter than I am, would think of that idea.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah well.. I guess back to the business of learning math...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hence the proposed. E=m(d/(t+x))^2

perl
 one year ago
Best ResponseYou've already chosen the best response.2Interesting idea, unfortunately this is outside my scope of knowledge. I will tag others :) @Michele_Laino

perl
 one year ago
Best ResponseYou've already chosen the best response.2Thinking outside the box is a good skill.

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0Yes. It describes what we observe.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0your formula: E=m(d/(t+x))^2 was correct if we deal with Classical Mechanics. When we have to deal with Relativistic Mechanics, we have to do another reasoning. The starting point is that the speed of a particle, with respect to a inertial system, can not be greater or equal to speed light c When a constant force is acting on a particle, its speed can not reach the constant C (speed light), so where it ends the extra energy that our particle receives from the work done by that force? The answer is: that extra energy which our particle receives from that external force, will transform itself in extra mass of our particle. In other words as the speed of our particle approaches to C, its inertia increases, according to this relation ship: \[\Large m = \frac{{{m_0}}}{{\sqrt {1  \frac{{{v^2}}}{{{c^2}}}} }} = \gamma {m_0}\] where m_0 is the mass of our particle, measured in a inertial system in which it is at rest.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0from such reasoning we are led to the correct meaning of the famous Einstein's relationship: \[\Large E = \Delta m \times {c^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome .. thanks for sharing.
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